Question

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field between the plates is E~ = 3.20 × 105 N/C. When the space is filled with a dielectric, the electric field is E~ = 2.50 × 105 N/C. What is the dielectric constant of this dielectric?

Answer 1

Answer:

The dielectric constant of this dielectric is 1.28.      

Explanation:

Given that,

When the space between the plates is evacuated, the electric field between the plates, $E_o=3.2\times 10^5\ N/C$

When the space is filled with a dielectric, the electric field is, $E=2.5\times 10^5\ N/C$

Let k is the dielectric constant of this dielectric. The electric field gets decreased by a factor of k if a dielectric is inserted between plates. So,

$k=\dfrac{E_o}{E}\\\\k=\dfrac{3.2\times 10^5}{2.5\times 10^5}\\\\k=1.28$

So, the dielectric constant of this dielectric is 1.28. Hence, this is the required solution.

Answer 2

Answer:

1.28

Explanation:

Electric field between the plates when the separation between the plates is air, Eo = 3.20 x 10^5 N/C

Electric field between the plates when the separation between the plates is dielectric, E = 2.5 x 10^5 N/C

Let K is the dielectric constant

By the definition of dielectric constant, it is the ratio of electric field in the vacuum to the electric field when the dielectric is placed.

$K =\frac{E_{0}}{E}$

$K = \frac{3.20\times 10^{5}}{2.5\times 10^{5}}$

K = 1.28

Thus, the dielectric constant of the material between the plates is 1.28.